00:01
So the compartment temperature, we have minus 70 degrees celsius, right? and exhausts into the room at 25 degrees celsius.
00:14
So we need to figure out how much work is done to change 0 .65 kilogram at 25 degrees celsius at minus water, right? into ice at this temperature, right? and then if the compressor has 105 watt output power and runs at 25 percent of all time, how long it will take, okay? so remember the work done by the refrigerator should be equal to k of k, right? we are looking at coefficient of performance here.
00:53
And coefficient of performance here, we're looking at tc over th minus tc, you know.
01:00
We're looking at these two temperatures of water and ice respectively, okay? so 25 will be th, which is basically 25 plus 273.
01:14
And this should get 301 kelvin.
01:17
The other one should be minus 17 plus 273.
01:23
This should be 298 kelvin.
01:31
273 and 17, you should get 256 k, okay? so therefore, the value of k should be equal to 256, okay? over 298 minus 256.
01:53
And you have the value of 6 .09, okay? let's say 6 .1.
02:01
6 .1 should be the value there, okay? so we have 0 .65 as the mass of water, okay? so therefore, our q will be equal to mass water, capacity water, changing temperature 1, plus mass water, latent heat of fusion, plus of ice, mass water, capacity of ice, and changing temperature 2, right? so therefore, we're looking at 0 .65 multiply 4186, multiply by 25, plus 0 .65 multiply 3 .3.
02:46
That is the latent heat of fusion, into 10 to minus 5, plus 0 .65 multiplied by 100, multiplied by 17...